Question

Joana solved the equation 4x^2−4x−3=0 . She followed the steps to solve a quadratic equation by grouping. However, in Step 3 she noticed that the two groupings did not share a common binomial, as shown in her work below. She concluded that the quadratic equation 4x^2−4x−3=0 has no solutions. Is Joana’s conclusion correct? Review Joana’s work to answer the question. Step 1: 4x^2+2x+(−6x)−3=0 Step 2: (4x^2+2x)+(−6x−3)=0 Step 3: x(4x+2)+(−3(2x+1))=0 (1 point) Responses No, because she did not factor out the GCF from the first grouping. No, because she did not factor out the GCF from the first grouping. No, because she can still continue the steps even if the two groupings have no shared binomial. No, because she can still continue the steps even if the two groupings have no shared binomial. Yes, because the two groupings do not share a common binomial. Yes, because the two groupings do not share a common binomial. Yes, because there are no factors of −12 that add up to −4 .

Answer 1

To determine if Joana's conclusion is correct, let's analyze her steps towards solving the quadratic equation $4x^2-4x-3=0$.

Step by step, Joana attempted to use the method of grouping by first rewriting the equation:

1. Step 1: Joana rewrote the equation as $4x^2+2x-6x-3=0$.

2. Step 2: She grouped the terms as $(4x^2+2x)+(-6x-3)=0$.

3. Step 3: In this step, she factored from the first grouping, obtaining $x(4x+2)-3(2x+1)=0$.

At this point, Joana is trying to factor, but she incorrectly believes that the two groupings do not share a common binomial. In her expression, $4x+2$ and $-3(2x+1)$ do not share a common binomial factor, which is correct. However, she can still proceed with solving the equation without a common binomial factor.

Evaluating Joana's Conclusion:

1. The fact that the groupings do not share a common binomial does not necessarily mean the original equation has no solutions. The equation could be factorable or solvable by other methods, like the quadratic formula.

2. To check for solutions, we can also use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, given $a=4$, $b=-4$, and $c=-3$:

   $$x=\frac{-(-4)\pm \sqrt{(-4{)}^2-4\cdot 4\cdot (-3)}}{2\cdot 4}$$ $$=\frac{4\pm \sqrt{16+48}}{8}$$ $$=\frac{4\pm \sqrt{64}}{8}$$ $$=\frac{4\pm 8}{8}$$

   This yields two potential solutions:

   • $x=\frac{12}{8}=\frac{3}{2}$
   • $x=\frac{-4}{8}=-\frac{1}{2}$

Since this shows that the equation has solutions, Joana's conclusion that there are no solutions is incorrect.

Final Answer: No, because she can still continue the steps even if the two groupings have no shared binomial.